[PPT] - Lecture 8: Finite State Machines And Sequential circuit Design PowerPoint Presentation

SLIDE 1

Lecture 8: Finite State Machines

And Sequential circuit Design

CSE 140: Components and Design Techniques for Digital Systems

Diba Mirza

Dept. of Computer Science and Engineering

University of California, San Diego

1

SLIDE 2

2

Generalized Model of Sequential Circuits

S(t) X Y CLK

SLIDE 3

3

Mealy Machine: yi(t) = fi(X(t), S(t)) Moore Machine: yi(t) = fi(S(t)) si(t+1) = gi(X(t), S(t)) C1 C2

CLK x(t) y(t)

Mealy Machine

S(t)

C1 C2

CLK x(t) y(t)

Moore Machine

S(t)

Canonical Form: Mealy and Moore Machines

SLIDE 4

4

C1 C2

CLK x(t) y(t)

Mealy Machine C1 C2

CLK x(t) y(t)

Moore Machine

S(t) S(t)

Differences in State Diagram: Mealy vs. Moore Machines

SLIDE 5

This Counter Design Is:

A. Moore machine
B. Mealy machine
C. None of the above

D Q D Q D Q OUT1 OUT2 OUT3 CLK "1"

SLIDE 6

Life on Mars?

6

This pattern recognizer should have

A. One state because it has one output
B. One state because it has one input
C. Two states because the input can be 0 or 1
D. More than two states because ….
E. None of the above

Mars rover has a binary input x. When it receives the input sequence x(t-2, t) = 001 from its life detection sensors, it means that the it has detected life on Mars J and the output y(t) = 1, otherwise y(t) = 0 (no life on Mars L). Implement the Life-on-Mars Pattern Recognizer!

SLIDE 7

Life on Mars?

7

Mars rover has a binary input x. When it receives the input sequence x(t-2, t) = 001 from its life detection sensors, it means that the it has detected life on Mars J and the output y(t) = 1, otherwise y(t) = 0 (no life on Mars L).

SLIDE 8

Mars Life Recognizer FSM

8

S1 S0

0/0 1/0 0/0 1/1

S2

0/0 1/0

Which of the following diagrams is a correct Mealy solution for the 001 pattern recognizer on the Mars rover?

A.

S1 S0

0/0 1/0 1/0 0/0

S2

1/1 B. 0/0

C. Both A and B are correct
D. None of the above

SLIDE 9

9

Pattern Recognizer ‘001’

S1 S0

0/0 1/0 0/0 1/1

S2

0/0 1/0

C1 C2

CLK x(t) y(t)

Mealy Machine

S(t)

What does state table need to show to design controls of C1?

A. (current input x(t), current state S(t) vs. next state, S(t+1))
B. (current input, current state vs. current output y(t))
C. (current input, current state vs. current output, next state)
D. None of the above

Mars Life Recognizer FFs

SLIDE 10

10

State Diagram => State Table with State Assignment

State Assignment S0: 00 S1: 01 S2: 10

S(t)\x 1 S0 S1,0 S0,0 S1 S2,0 S0,0 S2 S2,0 S0,1

S(t)\x 1 00 01,0 00,0 01 10,0 00,0 10 10,0 00,1 Q1(t+1)Q0(t+1), y

C1 C2

CLK x(t) y(t)

Mealy Machine

S(t)

S1 S0

0/0 1/0 0/0 1/1

S2

0/0 1/0

SLIDE 11

11

State Diagram => State Table => Excitation Table => Circuit

Q1(t) Q0(t)\x 0 1 00 01,0 00,0 01 10,0 00,0 10 10,0 00,1

id Q1Q0x D1 D0 y 000 1 1 001 2 010 1 3 011 4 100 1 5 101 1 6 110 X X X 7 111 X X X C1 C2

CLK x(t) y(t)

Mealy Machine

S(t)

SLIDE 12

12

0 2 6 4 1 3 7 5

x(t) Q1

0 1 X 1 0 0 X 0

Q0

D1(t): D1(t) = x’Q0 + x’Q1 D0 (t)= Q’1Q’0 x’ y= Q1x

State Diagram => State Table => Excitation Table => Circuit

id Q1Q0x D1 D0 y 000 1 1 001 2 010 1 3 011 4 100 1 5 101 1 6 110 X X X 7 111 X X X

SLIDE 13

13

D1(t) = x’Q0 + x’Q1 D0 (t)= Q’1Q’0 x’ y= Q1x

D Q Q’ D Q Q’

Q1 Q0 D1 D0 Q0 Q1 x’ x y Q’1 Q’0 x’

State Diagram => State Table => Excitation Table => Circuit

C1 C2

CLK x(t) y(t)

Mealy Machine

S(t)

SLIDE 14

Moore FSM for the Mars Life Recognizer

14

S1 S0

0/0 1/0 0/0 1/1

S2

0/0 1/0 Which of the following diagrams is a correct Moore solution to the ‘001’ pattern recognizer? A. B.

C. Both A and B are correct
D. None of the above

S1 S0

1 1

S2

1

S3 1

1

SLIDE 15

Moore Mars Life Recognizer: FF Input Specs

15

Pattern Recognizer ‘001’

What does state table need to show to design controls of C2?

A. (current input x(t), current state S(t) vs. next state, S(t+1))
B. (current input, current state vs. current output y(t))
C. (current state vs. current output y(t) and next state)
D. (current state vs. current output y(t) )
E. None of the above

C1 C2

CLK x(t) y(t)

Moore Machine

S(t)

S1 S0

1 1

S2

1

S3 1

1

SLIDE 16

Q1Q0\x 1 00 01,0 00,0 01 10,0 00,0 10 10,0 11,0 11 01,1 00,1 Q1(t+1)Q0(t+1), y

ID Q1Q0x D1 D0 y 000 1 1 001 2 010 1 3 011 4 100 1 5 101 1 1 6 110 1 1 7 111 1

S(t)\x 1 S0 S1,0 S0,0 S1 S2,0 S0,0 S2 S2,0 S3,0 S3 S1,1 S0,1

Moore Mars Life Recognizer: State Table

S1 S0

1 1

S2

1

S3 1

1

SLIDE 17

id Q1Q0x D1 D0 y 000 1 1 001 2 010 1 3 011 4 100 1 5 101 1 1 6 110 1 1 7 111 1

0 2 6 4 1 3 7 5

x(t) Q1

1 0 1 0 0 0 0 1

Q0

D0(t):

0 2 6 4 1 3 7 5

x(t) Q1

0 1 0 1 0 0 0 1

Q0

D1(t):

0 2 6 4 1 3 7 5

x(t) Q1

0 0 1 0 0 0 1 0

Q0

y(t):

Mars Life Recognizer: Combinational Circuit Design

SLIDE 18

Mars Life Recognizer Circuit Implementation

18

D Q Q’ D Q Q’

Q1 Q0 D1 D0 y

State Diagram => State Table => Excitation Table => Circuit

C1 C2

CLK x(t) y(t)

Moore Machine

S(t) D1(t)= Q1(t)Q0(t)’+Q1(t)’Q0(t) x(t) D0(t)= Q1(t)’Q0(t)’x(t)’+ Q1(t)Q0(t) x(t)’+Q1(t)Q0(t)’ x(t) y(t)= Q1(t)Q0(t)

SLIDE 19

FSM Specification

19

Q0(t) Q1(t)

D Q Q’ D Q Q’

CLK

x(t) Q0(t) Q1(t)

y(t)

SLIDE 20

From circuit to FSM

20

Q0(t) Q1(t)

D Q Q’ D Q Q’

CLK

x(t) Q0(t) Q1(t)

y(t) y(t) = Q1(t)Q0(t) Q0(t+1) = D0(t) = x(t)’ Q0(t)’ Q1(t+1) = D1(t) = x(t)’(Q0(t) + Q1(t))

SLIDE 21

21

State table

0 0 0 1 1 0 1 1

PS

input

x=0 x=1

Q1(t) Q0(t) | (Q1(t+1) Q0(t+1), y(t)) Present State | Next State, Output

Netlist ó State Table ó State Diagram ó Input Output Relation

Characteristic Expression:

y(t) = Q1(t)Q0(t) Q0(t+1) = D0(t) = x(t)’ Q0(t)’ Q1(t+1) = D1(t) = x(t)’(Q0(t) + Q1(t))

SLIDE 22

22

State table

0 0 0 1 1 0 1 1

PS

input

x=0 x=1 01, 0 00, 0 10, 0 00, 0 11, 0 00, 0 00, 1 00, 1

Q1(t) Q0(t) | Q1(t+1) Q0(t+1), y(t) Present State | Next State, Output

S0 S1 S2 S3

PS

input

x=0 x=1

S1, 0 S0, 0 S2, 0 S0, 0 S3, 0 S0, 0 S0, 1 S0, 1

Let: S0 = 00 S1 = 01 S2 = 10 S3 = 11

Remake the state table using symbols instead of binary code , e.g. ’00’ Netlist ó State Table ó State Diagram ó Input Output Relation

State Assignment

y(t) = Q1(t)Q0(t) Q0(t+1) = D0(t) = x(t)’ Q0(t)’ Q1(t+1) = D1(t) = x(t)’(Q0(t) + Q1(t))

SLIDE 23

23

Netlist ó State Table ó State Diagram ó Input Output Relation

S1 S2 S3 S0 S0 S1 S2 S3

PS

input

x=0 x=1

S1, 0 S0, 0 S2, 0 S0, 0 S3, 0 S0, 0 S0, 1 S0, 1

SLIDE 24

24

Netlist ó State Table ó State Diagram ó Input Output Relation Example: Given inputs and initial state, derive

utput sequence

Time 1 2 3 4 5 Input 1

State

S0 S1 S0 S1 S2 S3 Output 0 1

(0 or 1)

S0 S1 S2 S3

PS

input

x=0 x=1

S1, 0 S0, 0 S2, 0 S0, 0 S3, 0 S0, 0 S0, 1 S0, 1

x/y

S1 S2 S3

1

S0

1 1 1

SLIDE 25

25

Finite State Machine Example

Traffic light controller

– Traffic sensors: TA, TB (TRUE when there’s traffic) – Lights: LA, LB

TA LA TA LB TB TB LA LB

Academic Ave. Bravado Blvd. Dorms Fields Dining Hall Labs

SLIDE 26

26

FSM Black Box

Inputs: CLK, Reset, TA, TB
Outputs: LA, LB

TA TB LA LB CLK Reset Traffic Light Controller

SLIDE 27

27

FSM State Transition Diagram

Moore FSM: outputs labeled in each state
States: Circles
Transitions: Arcs

S0 LA: green LB: red Reset

TA LA TA LB TB TB LA LB

Academic Ave. Bravado Blvd. Dorms Fields Dining Hall Labs

SLIDE 28

28

FSM State Transition Diagram

Moore FSM: outputs labeled in each state

S0 LA: green LB: red S1 LA: yellow LB: red S3 LA: red LB: yellow S2 LA: red LB: green TA TA TB TB Reset Which of the following is true about the controller? A. The traffic light on Academic Ave (LA) remains green as long as there is traffic on that street B. The traffic light on both avenues are green for exactly once clock cycle in every four clock cycles

TA LA TA LB TB TB LA LB

Academic Ave. Bravado Blvd. Dorms Fields Dining Hall Labs

SLIDE 29

29

FSM State Transition Table

PS Inputs NS Output TA TB LA LB S0 X S1 S0 1 X S0 S1 X X S2 S2 X S3 S2 X 1 S2 S3 X X S0

S0 LA: green LB: red S1 LA: yellow LB: red S3 LA: red LB: yellow S2 LA: red LB: green TA TA TB TB Reset

SLIDE 30

30

FSM State Transition Table

PS Inputs NS Output TA TB LA LB S0 X S1 green red S0 1 X S0 green red S1 X X S2 yellow red S2 X S3 red green S2 X 1 S2 red green S3 X X S0 red yellow

S0 LA: green LB: red S1 LA: yellow LB: red S3 LA: red LB: yellow S2 LA: red LB: green TA TA TB TB Reset

State Encoding S0 00 S1 01 S2 10 S3 11 PS Inputs NS Q1(t) Q0(t) TA TB Q1(t +1) Q0(t +1) X 1 1 X 1 X X 1 1 X 1 1 1 X 1 1 1 1 X X

SLIDE 31

31

State Transition Table

PS Inputs NS Q1(t) Q0(t) TA TB Q1(t +1) Q0(t +1) X 1 1 X 1 X X 1 1 X 1 1 1 X 1 1 1 1 X X Q1(t+1)= Q1(t) xor Q0(t) Q0(t+1)= Q’1(t)Q’0(t)T’A + Q1(t)Q’0(t)T’B

SLIDE 32

32

FSM Schematic: State Register

S1 S0 S'1 S'0 CLK

state register

Reset r

SLIDE 33

33

Logic Diagram

S1 S0 S'1 S'0 CLK

next state logic state register

Reset TA TB

inputs

S1 S0 r

Q1(t+1)= Q1(t) xor Q0(t) Q0(t+1)= Q’1(t)Q’0(t)T’A + Q1(t)Q’0(t)T’B

S1=Q1 S0=Q0

SLIDE 34

34

FSM Output Table

PS Outputs Q1 Q0 LA1 LA0 LB1 LB0 1 1 1 1 1 1 1 1 1 1

Output Encoding green 00 yellow 01 red 10

LA1 = Q1 LA0 = Q’1Q0 LB1 = Q’1 LB0 = Q1Q0

PS Inputs NS Output TA TB LA LB S0 X S1 green red S0 1 X S0 green red S1 X X S2 yellow red S2 X S3 red green S2 X 1 S2 red green S3 X X S0 red yellow

SLIDE 35

35

FSM Schematic: Output Logic

S1 S0 S'1 S'0 CLK

next state logic

utput logic

state register

Reset LA1 LB1 LB0 LA0 TA TB

inputs

utputs

S1 S0 r

LA1 = Q1 LA0 = Q’1Q0 LB1 = Q’1 LB0 = Q1Q0

SLIDE 36

Summary: Implementation

36

Set up canonical form
Mealy or Moore machine
Identify the next states
state diagram ⇨ ¡state table
state assignment
Derive excitation table
Inputs of flip flops
Design the combinational logic
don’t care set utilization